If x=x(t) is the solution of the differential equation (t+1) d x=(2 x+(t+1)⁴) dt, x(0)=2, then, x(1) equals _________.

Solution

Maths sir · 4 steps

Divide by dt and it is a linear equation in x; the factor 1/(t+1)² does the rest.

1. Standard linear form. Divide both sides by (t+1) dt to get dx/dt - 2/(t+1)x = (t+1)³. This is linear in x with P = -2/(t+1) and Q = (t+1)³.
2. Integrating factor. ∫ P dt = -2 ln(t+1), so the integrating factor is e⁻²ˡⁿ⁽ᵗ⁺¹⁾ = 1/(t+1)². Multiplying through turns the left side into an exact derivative, and the equation separates.

   d/dt(x/(t+1)²) = t+1 ⇒ x/(t+1)² = (t+1)²/2 + C

3. Find C. Put t = 0, x = 2: 2/1 = 1/2 + C, so C = 3/2.
4. So. At t = 1: x/4 = 4/2 + 3/2 = 7/2, so x(1) = 4 × 7/2 = 14.

The common slip: 16. Taking C = 2 straight from x(0) = 2, without subtracting the 1/2, gives x/4 = 4 and so 16.

The idea to keep. For dx/dt + P x = Q, multiply by e^(∫ P dt) and integrate. Fix the constant from the initial condition before you substitute the new value.

Answer from the NTA final answer key. Solution by PrepWiser.